Convert your difference equation into differential equation:
y(i+1) = y(i)-y(i)*h;
y(i+1) - y(i) = -y(i)*h;
(y(i+1) - y(i))/h = -y(i);
dy/dt = -y
Time step h is replaced by dt to make the equation more understandable. Now, solve this differential equation using variable separation method:
dy/y = -dt
ln|y| = -t + C
y = exp(-t+C)
y = A*exp(-t)
At t = 0, A = y(0) = 3, in your case, y(0) is equivalent to y(1) in Matlab
y = 3*exp(-t)
This means that y is exponential. However, when solving using difference equation, make sure that h or the time is small enough to keep the solution stable. For this particular equation, h<1 . Use a larger value of h in your code just to see the effect.
